Find the velocity field of ideal fluid that has a sink strength of $2\pi k$ using complex analysis.

Question

The precise question is, find the velocity field of ideal fluid given that the fluid has a sink of strength $2\pi k$ at the origin (and no other singularities) and that it has velocity $V_\infty$ (a complex number) at infinity.

This is all the given information. I'm not really to sure where to begin. Any pointers would be appreciated! Thanks!

EDIT This is what I've come up with:

We have $$ \phi(z)=-\frac{2\pi k}{|z|}=\phi(x,y)=-\frac{2\pi k}{\sqrt{x^2+y^2}}, $$ which gives the velocity field $$ v(z)=v(x,y)=\nabla\phi(x,y)=\frac{2\pi kx}{(x^2+y^2)^{3/2}}+i\frac{2\pi ky}{(x^2+y^2)^{3/2}} $$ Now, as $z\to\infty$ we have $v\to0$. However, at infinity we must have $$ \lim_{z\to\infty}v(z)\equiv V_\infty $$ So, the velocity field becomes $$ v(x,y)=\frac{2\pi kx}{(x^2+y^2)^{3/2}}+i\frac{2\pi ky}{(x^2+y^2)^{3/2}}+V_\infty=\frac{2\pi k}{(x^2+y^2)^{3/2}}(x+iy)+V_\infty, $$ which gives the result.

Answer 1

From what I know about ideal fluids, the potential should be a harmonic function, which $1/|z|$ is not. I think the correct potential is $$\phi(z) = -k\log |z|$$ The constant $2\pi$ appears naturally: $\Delta \phi$ is $ - 2\pi k\delta $ where $\delta$ is the Dirac delta at $0$.

The real gradient can be taken using complex notation: $$ \nabla \phi = \frac{\partial}{\partial \bar z} (2\phi) = \frac{\partial}{\partial \bar z} (-k\log|z|^2 ) = \frac{\partial}{\partial \bar z} (-k\log z - k\log \bar z ) = -\frac{k}{\bar z} = -\frac{kz}{|z|^2}$$ Add $V_\infty$ to get the right behavior at $\infty$, as you did.

Answer 2

This is what I find in my (40 years old) notes: $$ \oint v_r\,ds = \int_0^{2\pi} v_r\,r\,d\theta = v_r\,r\,2\pi = -Q = -2\pi k \qquad \Longrightarrow \qquad v_r = -|u+i\,v| = -\frac{k}{r} $$ So, for that sink, when augmented with the velocity field at infinity: $$ u + i\,v = - \frac{k}{r} e^{i\,\theta} + V_\infty = - \frac{k}{r^2} \left[\,r \cos(\theta) + i\,r\sin(\theta)\,\right] + V_\infty = - \frac{k\,(x+i\,y)}{x^2+y^2} + V_\infty $$ Furthermore: $$ u - i\,v = -\frac{k}{r} e^{-i\,\theta} + \overline{V_\infty} = -\frac{k}{z} + \overline{V_\infty} = \frac{d\phi}{dz} $$ So the complex potential $\phi$ is (still apart from a complex constant): $$ \phi = - k\,\ln(z) + \overline{V_\infty}\, z $$ Which is all a bit different from the other answers.